There's an extra detail I'd like to mention
regarding the multiplicity of a zero and the graph of the polynomial:
You can tell from the graph whether an odd-multiplicity zero occurs only
once or if it occurs more than once.

What is the multiplicity
of x
= 5, given that
the graph shows a fifth-degree polynomial with all real-number roots,
and the root x
= –5 has a multiplicity
of 2?

The intercept at x
= –5 is of multiplicity
2.
The polynomial is of degree 5,
so the zero at x
= 5, the only other
zero, must use up the rest of the multiplicities. Since 5
– 2 = 3, then

x
= 5 must
be of multiplicity 3.

The zero at x
= 5 had to be of
odd multiplicity, since the graph went through the x-axis.
But the graph flexed a bit (the "flexing" being that bendy part
of the graph) right in the area of x
= 5. This flexing is
what tells you that the multiplicity of x
= 5 had to be more
than just 1.
In this particular case, the multiplicity couldn't have been 5
or 7
or more, because the degree of the whole polynomial was only 5,but
the multiplicity certainly had to be more than just 1.
Keep this in mind: Any odd-multiplicity zero that flexes at the crossing
point, like this graph did at x
= 5, is of multiplicity
3 or
more.

Which of the following
graphs could represent the polynomialf(x)
= a(x – b)2(x
– c)3?

Whatever this polynomial
is, it is of degree 5.
(I know this by adding the degrees on the two repeated factors: if I
multiplied everything out, the degree on the leading term would be 5.)

Since the polynomial is
of odd degree, Graph A can't be correct, because its ends
both go the same direction, meaning it is an even-degree polynomial.

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Since my polynomial has
two real-number zeroes (namely, zeroes at x
= b and at x
= c), I know
that Graph C can't be right: it only crosses the x-axis
once. So Graph C may be of odd degree, but it doesn't have enough zeroes.

From the end behavior,
I can see that Graph D is of odd degree. Also, I know that the negative
zero has an even multiplicity because the graph just touches the axis;
this zero could correspond to x
= b. But there
is no flexing where the graph crosses the positive x-axis,
so the odd zero here must have a multiplicity of only 1,
and I need the multiplicity of this zero to be more than just 1.
So Graph D might have the right overall degree (if the zero x
= b is of multiplicity
1),
but the multiplicities of the two zeroes don't match up with what I
need.

On the other hand, the
ends of the graph tell me that Graph B is of odd degree, and the way
the graph touches or crosses the x-axis
at the two graphed x-intercepts
tells me that the polynomial being graphed has one even-multiplicity
zero and one odd-and-more-than-1-multiplicity
zero. This matches what I need.

The correct graph
is Graph B.

Find the degree-7
polynomial corresponding to the following graph, given that one of the
zeroes has multiplicity 3.

From the graph, I can
see that there are zeroes of even multiplicity at x
= –4 and x
= 4. The zero at x
= –1 must be the zero
of multiplicity 3.
(This matches the graph, since the line goes through the axis, but flexes
as it does so, telling me that the multiplicity must be odd and must
be more than 1.)

Since the total degree
of the polynomial is 7,
and I already have multiplicities of 2,
2,
and 3 (which
adds up to 7),
then the zeroes at –4 and
4 must
be of multiplicity 2,
rather than multiplicity 4 or
multiplicity 6 or
something bigger.

They marked that one point
on the graph so that I can figure out the exact polynomial; that is,
so I can figure out the value of the leading coefficient "a".
Plugging in these x-
and y-values
from the point (1,
–2), I get:

a(1
+ 4)2(1 + 1)3(1 – 4)2 = a(25)(8)(9)
= 1800a = –2

Then a
= –1/900, and the polynomial,
in factored form, is:

y
= ( –1/900 )(x + 4)2(x
+ 1)3(x – 4)2

The exercise didn't say that
I had to multiply this out, so I'm not going to. The factored form (especially
for something as huge as this) should be a completely acceptable form
of the answer.